To solve this we must substitute #color(red)(4)# for #color(red)(a)# in the expression and then solve for #b#:
#5/(color(red)(a) - b) = 2# becomes:
#5/(color(red)(4) - b) = 2#
#5/(4 - b) xx color(red)((4 - b)) = 2color(red)((4 - b))#
#5/color(red)(cancel(color(black)(4 - b))) xx cancel(color(red)((4 - b))) = (2 xx 4) - (2 xx b)#
#5 = 8 - 2b#
#5 - color(red)(8) = -color(red)(8) + 8 - 2b#
#-3 = 0 - 2b#
#-3 = -2b#
#(-3)/color(red)(-2) = (-2b)/color(red)(-2)#
#3/2 = (color(red)(cancel(color(black)(-2)))b)/cancel(color(red)(-2))#
#3/2 = b#
#b = 3/2#